Lottery, probability to win

Hello, I have problem with one probability theory task. Hope that someone of you will be able to help me.

So task is: Suppose that you are playing in lottery. The comptuer generates the lottery ticket which is made from 25 numbers. Total there are 75 numbers and 49 are extracted during the game. You win if all 25 numbers from 49 will be extracted. Need to calculate probability of winning.

So my soulution is very simple: (50;24)/(75;49). The 50 is amount of numbers which are not included in our ticket. 24 is 49-25.

The reasoning is that if we consider the 50 numbers that are not ours and the 75 total numbers, we win if and only if the 26 drawn from each set are the same.

What are the sets? By context, the sets are "the 50 numbers that are not ours" and "the 75 total numbers". But 26 drawn from each set are only the same if none of them are the numbers we picked ... and where do you get the number 26 from anyway: the numbers the lottery didn't pick?

Did Mathman get the game description correct?

I think that, in order for you to understand your result better, you need to describe it more carefully.

^
Yes the 26 numbers are those not drawn.
Suppose we win.
The picking process splits the 50 numbers that are not ours into 24 picked and 26 not.
The 75 numbers split into 49 picked and 26 not.
Every possible split is equally likely.
I don't think the OP description is not bad, Mathman and the OP's description seem in agreement. You pick 25 numbers, the lottery picks 49, you win if yours are a subset of the lotteries. I think it is easier to visualize if we imagine the lottery picking 26 and you win if the lottery picks none of yours.

The first number drawn causes us to lose with probability 1/3 which increases with each number drawn until the final draw which causes loss with probability 1/2. The total probability of winning is.
(50/75)(49/74)(48/73)...(50-k)/(75-k)...(23/52)(24/51)(25/50)
or
$$ \prod_{k=0}^{25} \frac{50-k}{75-k} = \frac{50!49!}{75!24!} = {50 \choose 26}/{75 \choose 26} $$